A parallelogram a micron wrong

A parallelogram linkage is the four-bar everybody has handled. Two equal cranks swing from two pivots, a coupler as long as the distance between the pivots joins their ends, and the coupler stays parallel to the ground at every angle. Drafting machines, desk lamps, the coupling rods between a locomotive’s driving wheels and the arms of a pantograph are all built on it, and its motion is the simplest there is: the output crank turns exactly as the input does.

Twice in every turn it does something less simple. When the input points straight along the ground line, all four bars lie in one line, and at that instant the chain can go on in two ways: as a parallelogram, or crossed, with the coupler turning over and the output crank running the opposite way. That configuration is a change point, and the usual account of it is about the choice: which way does the linkage go, and what stops it choosing wrong?

This essay asks a prior question. The choice exists only if the lengths are exactly right. A linkage that has been made has lengths that are not exactly anything, and the question is what the flat position looks like to a parallelogram that is a micron wrong.

Two walls through one linkage #

Eight kinds of four-bar sorted every four-bar by the signs of three sums of its lengths. With the ground g, the input a, the coupler b and the output c, the sums are $T_1 = g + c - a - b$, $T_2 = g + b - a - c$ and $T_3 = b + c - a - g$. Each is zero exactly when the four bars can be laid flat in one of three arrangements, and the planes where they vanish are the walls between regions, the only places where a four-bar’s motion can change its kind.

The parallelogram used throughout this essay has a ground of 3, two cranks of 2 and a coupler of 3, the proportions of the three-crank chain in the mechanism Grübler says cannot move. Its sums are $T_1 = 0$, $T_2 = 2$ and $T_3 = 0$. Two of the three vanish at once. The linkage is not beside a wall; it is on two, at the line where they cross, and four regions of length space meet there.

That is a statement about every parallelogram, not about these lengths. Adding and subtracting the two sums gives $T_1 + T_3 = 2(c - a)$ and $T_1 - T_3 = 2(g - b)$, so both vanish exactly when the two cranks are equal and the coupler equals the ground, which is the definition of a parallelogram. The corner is the whole family.

It follows that nothing made is at the corner. Change any single length by any amount δ and at least one of the two sums moves off zero, and the linkage is inside one of the four regions around it. There are eight single-length errors, each bar long or short, and their regions are measured, not guessed, by the closure test at 3,600 input angles.

The eight errors land in exactly four regions, two in each. An input that is short and an output that is long make a crank-rocker. An input that is long and an output that is short make a rocker-crank. A coupler that is long and a ground that is short make the π–0 triple rocker. A coupler that is short and a ground that is long make the 0–π triple rocker.

So the answer to “what is a parallelogram a micron wrong” is not “a parallelogram, nearly”. It is one of four different machines, and the sign of the error decides which one, while its size decides nothing about the kind. The pairing also says which errors a workshop has to control. Two cranks drilled together in one setup can both be a tenth of a millimetre long and the linkage is still a parallelogram, because only the differences c − a and g − b move the sums. What cannot be tolerated in either direction is a mismatch within a pair.

What an error does at the flat position #

Why four regions, and what each does, becomes clear by looking at the flat position closely.

With the input at angle θ and the output at angle ψ, both measured from the ground line, the chain closes when

$$F(\theta, \psi) = g^2 + a^2 + c^2 - b^2 + 2gc\cos\psi - 2ga\cos\theta - 2ac\cos(\psi - \theta) = 0.$$

For the parallelogram at θ = ψ = 0, F and both of its first derivatives vanish, which is what makes the flat position special: the equation has no slope there to decide which way the curve goes. What decides it is the second-order part,

$$Q(\theta, \psi) = a(g+a)\,\theta^2 - 2a^2\,\theta\psi - a(g-a)\,\psi^2,$$

and Q factorises into two straight lines through the origin. One is ψ = θ, the parallelogram. The other is ψ = −(g + a)θ/(g − a), which for these lengths is ψ = −5θ, the crossed linkage, whose output runs backwards five times as fast as the input near the flat position. The two motions cross there, and that crossing is the change point.

A length error adds a constant. Changing any one bar by δ changes F at the flat position by ±2gδ, the sign depending on which bar and which way. So near the flat position the configuration curve is Q = ∓2gδ, which is not two crossing lines but a hyperbola, and a hyperbola near its centre has only two ways to sit.

In one, it opens along the output axis. The input can pass through the flat position, but the two motions no longer meet: one branch arrives along the crossed line and leaves along the parallelogram’s, the other does the reverse, and between them at θ = 0 is a gap in the output angle. In the other, it opens along the input axis. Neither branch reaches θ = 0; each arrives, turns back and leaves, and the input has a stall, an interval around the flat position it can never enter.

A shorter input opens a gap; a longer coupler makes a stall. In the language of configuration space, a gap is the two crossing circuits reconnecting into two separate ones, and a stall is them reconnecting into one that folds back. The parallelogram has a second flat position, at θ = π, where the input points away from the other pivot, and the same expansion holds there with g − a and g + a exchanged. So every error has a verdict at each of the two flat positions, and the four regions are exactly the four combinations. A crank-rocker has a gap at both flat positions, and its input turns all the way round. A rocker-crank has a stall at both, and its input only swings. Each triple rocker has one of each, which is why its rocker’s swing contains one flat position and not the other. The expansion and the region census are two routes to the same four machines, and at the corner they agree on all eight errors.

A square root, measured over nine decades #

The hyperbola also says how large the gap and the stall are. Solving Q = ∓2gδ for the output angle gives a discriminant that, at the flat position, is proportional to δ, so both quantities are square roots of the error:

$$\theta_{\text{stall}} = \sqrt{\frac{2(g-a)\,\delta}{ag}}, \qquad \psi_{\text{gap}} = 2\sqrt{\frac{2g\,\delta}{a(g-a)}}.$$

For these lengths the stall is $\sqrt{\delta/3}$ and the gap is $2\sqrt{3\delta}$. That is a leading-order law, and it can be wrong in two ways: the expansion might have dropped a term that matters, or the constant might be miscalculated. So the gap and the stall are also measured with no expansion at all, and the two routes share nothing but the four lengths.

That is the discipline of two routes to a sensitivity, applied to a sensitivity that is not a derivative, since the square root has an infinite slope at nought. The measured gap is the difference between the two circle intersections that place the output pin, evaluated at the flat input angle itself. The measured stall is found by bisection on whether those two circles meet, starting from the flat angle and halving the interval four hundred times. Neither computation mentions a derivative, a hyperbola or a square root.

The dots sit on the lines. At δ = 10⁻¹⁰ the stall measures 5.773501 × 10⁻⁶ radians against a law of 5.773503 × 10⁻⁶, and the gap 3.464102 × 10⁻⁵ against 3.464102 × 10⁻⁵. The log-log slopes over every error up to 10⁻⁴ are 0.500001 and 0.499999. The law begins to fail only where it should, when the error is no longer small: at δ = 10⁻¹ the stall is 2.6 per cent above the law.

At the second flat position the constants change and the exponent does not. There a short input still opens a gap, of $2\sqrt{0.6\,\delta}$, and a short coupler stalls the input by $\sqrt{5\delta/3}$. Both measured slopes are 0.500000.

What a square root means is easiest to see at a size somebody might build. Scale the linkage so the ground is 300 mm, which makes the cranks 200 mm and turns an error of 10⁻⁵ into one micron. A coupler one micron too long stops the input 0.1046° short of the flat position, and a crank pin 200 mm from its pivot therefore loses 0.365 mm of travel it would have had. One micron of length has become 365 microns of motion. An input crank one micron short opens a gap of 0.628° in the output. The ratio between the error and its consequence is not fixed; it grows without limit as the error shrinks, because $\sqrt{\delta}/\delta$ does.

The output turns round #

The gap has a second consequence, and it is the one a machine would feel.

Follow the crank-rocker, the linkage with its input a little short, through the flat position on one of its two circuits. It arrives along the crossed line, with its output running backwards at five times the input’s speed, and leaves along the parallelogram’s line, with its output running forwards at the input’s speed. In between, the output’s speed passes through zero. The output stops and reverses, inside a window of input angle a few square roots of δ wide.

The speed never exceeds the five it started at, so nothing about the velocity is alarming, and nothing about the transmission angle warns of it either, since a parallelogram’s coupler lies along both cranks at the flat position whether or not the lengths are exact. The acceleration is where the error shows. The speed changes by six units of input angle per unit over a window of width about $\sqrt{\delta}$, so the output’s angular acceleration at the flat position grows as one over $\sqrt{\delta}$. Differentiating the hyperbola twice gives it exactly,

$$\psi''_{\max} = \frac{a\,g^2}{(g-a)\sqrt{2ag(g-a)\,\delta}},$$

which is $3\sqrt{3}/\sqrt{\delta}$ for these lengths. The second route is the exact closure again, differentiated implicitly at the flat position with no expansion: at δ = 10⁻⁵ it gives 1,643.13 per radian squared against a law of 1,643.17, and the log-log slope is −0.500000.

The right-hand panel is the leading-order law in its most compact form. Measured in units of $\sqrt{\delta}$, every such reversal is the same event: the three curves for errors a hundred times apart lie on one shape, and the dashed curve is that shape computed from the hyperbola. The size of the error sets only the scale of the window and the height of the peak.

On the 300 mm linkage with its input a micron short, running at 100 revolutions a minute, that peak is 1.8 × 10⁵ radians per second squared at the flat position. At the end of a 200 mm output crank it is 36,000 m/s², about 3,700 g, twice every turn. That number is a statement about a rigid linkage with perfect pins, and no real one would reach it: something would bend, or a clearance would open, first. What it shows is that a rigid machine of these proportions cannot run through its flat position, and that the obstacle grows as the machine is made more accurately.

Over a whole turn the consequence is stranger still. With a gap at both flat positions, each circuit follows the parallelogram for half a turn and the crossed linkage for the other half. The output rises through a little under 180° while the input turns half a revolution, and falls back through the same angle while the input turns the other half: 179.55° each way for the micron-short input. A parallelogram with a short input is a crank-rocker whose rocker swings nearly half a turn. Only the flat positions reveal it, and a drawing at any other angle cannot tell the two apart.

The third bar #

Almost every parallelogram that has to pass through its flat positions carries a third crank, parallel to the other two and pinned to the coupler at a third point.

The mechanism Grübler says cannot move is that chain, and its paradox is well known: five links and six pins give a count of zero, and it moves. The rank of the constraint Jacobian is five against six coordinates, so it has one freedom, because the third crank is redundant. The usual reason for adding it is the one this essay started from, the choice at the flat position. The crossed motion is not available to three cranks at once, since a second crossed linkage would need the third crank to turn the other way from the first, so a three-crank chain carries straight through as a parallelogram.

That reason is correct for the exact linkage. For a built one there is a better one. A three-crank chain cannot spend an error on a gap or a stall, because both are ways of leaving the parallelogram’s motion near the flat position, and it has no other motion to leave for. So the error has to go somewhere else, and where it goes can be measured: follow the exact parallelogram motion with one crank short by δ and ask how far any bar would have to stretch to stay assembled.

The coupler between the short crank and its neighbour would have to change length by δ·|cos θ|, which is largest, at exactly δ, at the flat positions themselves. With the coupler long instead, the misfit is δ at every angle. Measured over 3,600 angles for errors from 10⁻⁸ to 10⁻², the largest misfit is δ to six significant figures for both.

That is the third crank’s other job, and it changes the order of the error rather than its size. Without it, an error of δ becomes a stall or a gap proportional to $\sqrt{\delta}$ and a reversal whose acceleration grows as $1/\sqrt{\delta}$. With it, the same error becomes a demand for δ of clearance in some pin, or δ of stretch in some bar, and a pin clearance of a micron is an ordinary thing to have. On the 300 mm frame the choice is between 365 microns of lost travel and one micron of play.

The price is the one every redundant constraint charges, and it is paid in the opposite direction. A three-crank chain with a mismatched crank and no clearance at all is exactly what Grübler’s count said it was: a structure, which will not turn. The redundancy that removes the square root is the redundancy that makes a tolerance compulsory, as a seventh contact does for a part and a micron does for a compiled straight line. The difference here is that the compulsory tolerance is linear in the error and the sensitivity it replaces is not.

What the model leaves out #

Clearance in the two-crank chain. Every figure above has perfect pins. A real parallelogram with a little play at each pin can reach configurations a rigid one cannot, because a clearance behaves like an extra short link, and whether it can cross the gap between its two circuits depends on how the gap in output angle compares with the play. That comparison is not made here.

Errors in more than one bar. Only single-length errors are measured. Because only c − a and g − b move the sums, two errors that match, such as both cranks short by the same amount, leave the linkage a parallelogram of a different size; two that do not match land in one of the four regions by the signs of those two differences.

Elasticity and dynamics. The acceleration at the flat position is a kinematic quantity for a rigid chain. How much of it a real linkage transmits, and what the bars and pins do instead, is a question about stiffness and mass that this essay does not answer.

How real machines avoid the flat position. A locomotive couples the two sides of its driving wheels with cranks set 90° apart, so that one side is never at its flat position when the other is, and a drafting machine often uses two parallelograms out of phase. Both are ways of never needing the change point, and neither is analysed here.

What comes next: the play that gives the choice back #

The comparison left open above has a definite form. At the flat position, the two circuits of a parallelogram with its input short by δ are $2\sqrt{3\delta}$ apart in output angle, but the length change needed to join them is only of order δ, because the exact parallelogram, with its crossing, is a length change of δ away. So a pin clearance of roughly δ should restore the change point, and the choice with it, while leaving the square-root gap in place for every clearance smaller than that.

Its distinct argument would be a measurement of that threshold: the smallest radial clearance, at one pin or shared among the four, at which a built parallelogram can pass from one circuit to the other at the flat position, as a function of δ and of which pin carries the play. If the threshold is exactly the length error, the result is that a parallelogram’s change point is restored by clearance at linear cost, which is the same order the third crank charges, arrived at without a third crank.